Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 4 - Section 4.8 - Solving Equations Containing Fractions - Exercise Set: 11

Answer

The solution is $\frac{11}{16}$

Work Step by Step

7z + $\frac{1}{16}$ - 6z = $\frac{3}{4}$ Simplify 7z-6z=z z +$\frac{1}{16}$ = $\frac{3}{4}$ Subtract $\frac{1}{16}$ to both sides z +$\frac{1}{16}$ - $\frac{1}{16}$ =$\frac{3}{4}$ - $\frac{1}{16}$ LCD of 4 and 16 is 16 z=$\frac{12}{16}$ - $\frac{1}{16}$ z=$\frac{12-1}{16}$ z=$\frac{11}{16}$ The solution is $\frac{11}{16}$ Check 7z +$\frac{1}{16}$ -6z=$\frac{3}{4}$ 7$\times$$\frac{11}{16}$ +$\frac{1}{16}$ -6$\times$$\frac{11}{16}$=$\frac{3}{4}$ $\frac{77}{16}$ +$\frac{1}{16}$ -$\frac{66}{16}$=$\frac{12}{16}$ $\frac{77+1-66}{16}$=$\frac{12}{16}$ $\frac{12}{16}$=$\frac{12}{16}$
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